# Integral involving exponential function

Can you give any hints how to solve the integral $$\int\limits_{0}^{\infty} {\exp \left( \frac{-w (z-u)^2}{2u^2 z} \right) dz}$$ for $u>0, w>0$, or does no close form exist? Substitution seems to be difficult…

• I would suggest expanding the square: $\frac{-w(z- u)^2}{2\mu^2z}= \frac{-wz^2+ 2wuz- wu^2}{2\mu z}= \frac{-w}{2\mu}z+ \frac{wu}{\mu}- \frac{wu^2}{2\mu}\frac{1}{z}$. The "1/z" is the hard part. – user247327 Jun 7 '17 at 19:45

## 2 Answers


$\ds{\,\mrm{K}_{\nu}}$ is a Modified Bessel Function. See A & S Table.

$$\bbx{\int_{0}^{\infty}\exp\pars{-\,{w\bracks{z - u}^{2} \over 2u^{2}z}}\,\dd z = 2u\expo{w/u}\,\mrm{K}_{1}\pars{w \over u}} \qquad\qquad \verts{\mrm{arg}\pars{w \over u}} < {\pi \over 2}$$

• @jana1802 Fixed. Thanks. It's funny yesterday it was right but last night I 'recheck' everything and made the mistake you pointed out. – Felix Marin Jun 9 '17 at 6:56

By expanding the square and performing a substitution we have that the given integral depends on a modified Bessel function of the second kind, since

$$\forall \alpha>0,\qquad \int_{0}^{+\infty}\exp\left(-\alpha\left(w+\frac{1}{w}\right)\right)\,dw = 2\cdot K_1(2\alpha).$$